TITLE: Is the infinite-dimensional unit sphere compact?
QUESTION [10 upvotes]: Riesz' lemma gives us that in infinite-dimensional spaces no ball is compact, but what about the sphere $\{x \in X : \|x\| = 1\}$? Can we say something about the compactness of the sphere in infinite-dimensional spaces?
(I guess the sphere is also not compact and I think one can also show this by constructing a sequence with Riesz lemma that has no convergent subsequence). Is this idea correct?

REPLY [22 votes]: Suppose that the unit sphere $S_X$ of $(X, \| . \|_X)$ is compact. 
Then the unit ball of $X$ is the image of the compact set $[0,1] \times S_X$ by the continuous map $(t, v) \mapsto tv$, and hence is compact.

REPLY [10 votes]: Sounds right to me. The sequence of points $(1,0,0,\ldots), (0, 1, 0, \ldots), (0, 0, 1, \ldots), \ldots$ is a sequence of points on the sphere that has no convergent subsequence, because the distance between any two of the points is $\sqrt{2}$.

REPLY [5 votes]: Hint: In a metric space, sequential compactness and compactness are equivalent. Now consider a sequence consisting of unit length basis vectors.